Calculus Question?

Question

Ok, for the function f(x)=(e^x)*cos(x), does it only have 1 point of inflection, at x = pi, in the interval [0,2 pi]?

thanks

Answer 1

Well let's take our function and get the second derivative.

f(x) = e^x * cos(x)
f'(x) = -e^x * sin(x) + e^x * cos(x)
f''(x) = -e^x * cos(x) - e^x * sin(x) - e^x * sin(x) + e^x * cos(x)

Simplifying our last equation leads us to:

f''(x) = -2e^x * sin(x)

We can see that our second derivate is only 0 if our sin(x) is equal to zero, since the exponential function always non-zero. sin(x) is zero at 0 + nPI, where n is any integer.

If we do a quick check for sign on our second derivate at these points within our interval of [0, 2PI] we get the following:

+ 0 - PI + 2PI

So we have points of inflection on the graph at 0, PI, and 2PI. However, our domain only contains the value 0 but nothing before, and it contains the value 2PI but nothing afterwards, and by definition, a point of inflection occurs in an open interval such as (a, b). Since our domain does not contain any possible open intervals that contain either 0 or 2PI, I would agree with you, that there is only one, at x = PI as you stated.

--charlie

Answer 2

To understand the answer, we must realize the following definitions:

On a given interval I,
f''(x) > 0 if the function is concave up and
f''(x) < 0 if the function is concave down, thus
If f''(x) = 0, or f''(x) does not exist, then it is very likely to be a point of inflection!

f''(x) = f'(f'(x))..= d/dx[d/dx(f(x))]

According to the product rule:
d/dx[f(x)g(x)] = f(x)g'(x) + g(x)f'(x)

So, d/dx[(e^x)*(cos(x))] = -(e^x)sin(x) + (e^x)cos(x) = G
and
d/dx(G) = f''(x) = -(e^x)cos(x) + -(e^x)sin(x) + -(e^x)sin(x) + (e^x)cos(x) = -2(e^x)sin(x)

When -2(e^x)sin(x) = 0, we have found our points of inflection.
For the increment of the x-axis from zero to 2pi this gives us three points of inflection (mathematically-speaking):
0, pi, and 2pi are the points of inflection {-2(e^x)sin(x) = 0} in the zone of 0 < or = x < or = 2pi.

Therefore, to repeat, the solution (which can be approximately double-checked visually on paper or graphing calculator) is:
{0, pi, 2pi} or {0, 3.1415..., 6.2831...} if we approximate...
If we exclude the end-points of the increment, then the solution is simply: {pi} or {3.1415...} approximately...

I hope this helped. Any need for clarification? Any desire to compliment or criticize the style or veracity of the solution statement? Any gratitude or criticisms?... Anything helpful or positive welcome.

Be well and god bless.

~Mathephiliac!

Answer 3

I've done this rather quickly so may have made a mistake.
I find two turning points in the given interval but neither is a point of inflection. One is a local maximum and the other a local minimum. Also neither is at x = pi.

Later edit. I have missed a P. of I. at x = pi but I would point out that second differential = 0 does not guarantee you have a
P. of I. Second differential can be zero at maxima and minima. F(x) = x^4 is an example.

Answer 4

as you know for finding the inflection point you should find the second derivative of the function;
at the inflection point second derivative of the function is zero or undefined;
in the case of your function Second derivative is
f"(x)= -2*(e^x) * sin(x); equating by zero we get
x = K*pi K=0,1,2,...
x --> -infinity
in the 0 to 2pi, pi is answer

Note: 0 & 2pi are also inflection point of function but not in this interval.

Answer 5

f(x) = (e^x)cos(x)

First off, points of inflection are determined by finding the second derivative and making it 0.

f'(x) = (e^x)cos(x) + (e^x)(-sin(x))
f'(x) = (e^x)cos(x) - (e^x)(sin(x))

Factoring out e^x,

f'(x) = [e^x] [cos(x) - sin(x)]

Solving for the second derivative,

f''(x) = [e^x] [cos(x) - sin(x)] + [e^x][-sin(x) - cos(x)]
f''(x) = [e^x] [cos(x) - sin(x)] - [e^x][sin(x) + cos(x)]
f''(x) = (e^x) [cos(x) - sin(x) - (sin(x) + cos(x))]
f''(x) = (e^x) [cos(x) - sin(x) - sin(x) - cos(x)]
f''(x) = (e^x) [-2sin(x)]

f''(x) = -2(e^x) sin(x)

To find the points of inflection, equate this to 0.

0 = -2(e^x)sin(x)
0 = (e^x)sin(x)

To find the solutions to this, equate each factor to 0.
e^x = 0
sin(x) = 0

But e^x has no solution, so we move onto sin(x) = 0.

sin(x) = 0; therefore, in the given interval stated,
x = {0, pi, 2pi}

Given that the graph cannot inflect before 0 and after 2pi (since the interval's restriction cannot make it happen), the only point of inflection is at x = pi.

Answer 6

it equals this: 2+2=5